What field does this Eigenvalue belong to? Suppose  $\mathbb{C}^n$ is a Vector Space over $\mathbb{R}$.
Given a Linear Operator T : $\mathbb{C}^n$ $\rightarrow$ $\mathbb{C}^n$, it follows that $\lambda$ is an Eigenvalue of  T if and only if there exists some nonzero vector x $\in$ $\mathbb{C}^n$  such that T(x) = $\lambda$x.
My question is, exactly what field does $\lambda$ belong to? Does $\lambda$ belong to $\mathbb{C}$ or $\mathbb{R}$?
 A: In addition to other good points made in comments and answer: yes, there is some ambiguity in the idea of "eigenvalue" for linear maps/operators on vector spaces over fields (such as $\mathbb R$) that are not algebraically closed. And we should not get distracted by the idea that $\mathbb C^n$ can be a vector space over $\mathbb R$... that is mostly a distraction, without genuine mathematical content.
For example, consider operator $T=\pmatrix{0&-1\cr 1&0}$ on $\mathbb R^2$. If we look at the characteristic equation $\lambda^2+1=0$, we see that there are no real eigenvalues. But $\pm i$ are eigenvalues... if we extend the vector space to its complexification...
(Choosing a basis also doesn't really have much to do with this phenomenon...)
A: If you look at $\mathbb C^n$ as a $\mathbb R$-vectorspace the eigenvalue $\lambda$ has to be from $\mathbb R$, thats how it is defined. If $T$ is $\mathbb R$-linear it doesn't even have to be $\mathbb C$-linear, so it wouldn't even be a well defined question.
If you then look at maps that are also $\mathbb C$-linear you can say that the map comes from multiplication with an element $(\phi,r)$ from $\mathbb C$ this then corresponds to turn by $\phi$ and dilation by $r$ of the complex "Eigenvector" corresponding to $a+bi$.
